Find the number of solutions of the equation: x + 2 tan x = π 2 , when x ∈ [ 0 , 2 π ]

Given equation is x + 2 tan x = π 2 tan x = π 4 - x 2 If we plot the graph of the curve y = tan x and y = π 4 - x 2 in the interval [ 0 , 2 π ] intersect at three points. Hence, number of solutions is three.

Find the number of solutions of the equation: x + 2 tan x = π 2 , when x ∈ [ 0 , 2 π ]

Given equation is x + 2 tan x = π 2 tan x = π 4 - x 2 If we plot the graph of the curve y = tan x and y = π 4 - x 2 in the interval [ 0 , 2 π ] intersect at three points. Hence, number of solutions is three.

Find the set of all x in ( - π , π ) satisfying | 4 sin x - 1 | < 5 .

Given equation | 4 sin x - 1 | < 5 ( now use x < a ⇒ - a < x < a ) ⇒ - 5 < 4 sin x - 1 < 5 ⇒ - 5 + 1 4 < sin x < 5 + 1 4 Principal argument for which sin x = - 5 + 1 4 is - π 10 . Principal argument for which sin x = 5 + 1 4 is 3 π 10 . Hence, it is clear from the graph that it has 3 intervals for the solution. x ∈ - π , - 9 π 10 ∪ - π 10 , 3 π 10 ∪ 7 π 10 , π

If x ∈ [ 0 , π ] and 81 sin 2 x + 81 cos 2 x = 30 , then find x .

Let 81 sin 2 x = y Then 81 cos 2 x = 81 1 - sin 2 x = 81 . 81 - sin 2 x = 81 · y - 1 Now the given question can be written as y 2 - 30 y + 81 = 0 Using factorization, we get ⇒ y = 3 or y = 27 ⇒ 81 sin 2 x = 3 or 27 ⇒ 3 4 sin 2 x = 3 or 3 3 ⇒ sin 2 x = 1 4 or 3 4 when x ∈ 0 , π ⇒ x = π 6 , π 3 , 5 π 6 , 2 π 3

If x ∈ [ 0 , π ] and 3 sin 2 x + 2 cos 2 x + 3 1 - sin 2 x + 2 sin 2 x = 28 , then find x .

We have 3 sin 2 x + 2 cos 2 x + 3 1 - sin 2 x + 2 sin 2 x = 28 or 3 sin 2 x + 2 cos 2 x + 3 3 - sin 2 x + 2 cos 2 x = 28 Let 3 sin 2 x + 2 cos 2 x = y Now given question can be written as y + 27 y = 28 ⇒ y 2 - 28 y + 27 = 0 y = 27 , 1 ⇒ 3 sin 2 x + 2 cos 2 x = 3 3 or ⇒ 3 sin 2 x + 2 cos 2 x = 3 0 ⇒ sin 2 x + 2 cos 2 x = 3 or sin 2 x + 2 cos 2 x = 0 ⇒ sin 2 x + cos 2 x = 2 or 2 cos x ( sin x + cos x ) = 0 ⇒ sin 2 x = cos 2 x = 1 or ( cos x = 0 or tan x = - 1 ) Which is not possible that sin 2 x = cos 2 x = 1 , for any value of x and so y ≠ 27 . ⇒ cos x = 0 or tan x = - 1 ⇒ x = π 2 , 3 π 4

Solve the equation: cos 3 x · cos 3 x + sin 3 x · sin 3 x = 0

We have cos 3 x · cos 3 x + sin 3 x · sin 3 x = 0 Using cos 3 x = 4 cos 3 x - 3 cos x , sin 3 x = - 4 sin 3 x + 3 sin x. 1 4 cos 2 3 x + 3 cos x cos 3 x + 3 sin x sin 3 x - sin 2 3 x = 0 ⇒ 1 4 [ 3 cos 2 x + cos 6 x ] = 0 ⇒ cos 3 2 x = 0 (Put 2 x = t and use cos 3 t + 3 cos t = 4 cos 3 t ) ⇒ 2 x = 2 n + 1 π 2 ⇒ x = 2 n + 1 π 4

Solve for x , which satisfies sin x + cos x = min a ∈ R 1 , a 2 - 4 a + 6

We know a 2 - 4 a + 6 = a - 2 2 + 2 > 1 for all a as it is 2 + (a positive quantity), So, min a ∈ R 1 , a 2 - 4 a + 6 = 1 ∴ the given equation can be written as- sin x + cos x = 1 Multiply the equation by 1 2 to both side ⇒ sin x + π 4 = sin π 4 ⇒ x = n π + - 1 n π 4 - π 4

Solve for α , which satisfy: max { 5 sin θ + 3 sin θ - α } = 7 , θ ∈ R

From the given question 5 sin θ + 3 sin θ cos α - cos θ sin α = 5 + 3 cos α sin θ - 3 sin α · cos θ By the concept of maximum value of a sin θ + b cos θ , max θ ∈ R 5 sin θ + 3 sin θ - α = 5 + 3 cos α 2 + 9 sin 2 α = 34 + 30 cos α ∴ According to question, 34 + 30 cos α = 49 ⇒ cos α = 1 2 ⇒ cos α = cos π 3 ⇒ α = 2 n π ± π 3

Solve: sin x + cos x - 2 2 sin x · cos x = 0

We have, sin x + cos x = 2 2 sin x cos x Use formula sin x cos θ + cos x sin θ = sin x + θ and 2 sin x cos x = sin 2 x , ⇒ 2 1 2 sin x + 1 2 cos x = 2 sin 2 x ⇒ sin x + π 4 = sin 2 x ⇒ 2 x = n π + - 1 n x + π 4 Taking n even; x = 2 m π + π 4 Taking n odd, x = 2 m + 1 3 π - π 12 Thus, x = 2 m + 1 4 π , 1 3 2 m + 3 4 π , where m ∈ Z .

Solve : sec θ - cosec θ = 4 3

We have sec θ - cosec θ = 4 3 sin θ - cos θ = 4 3 sin θ cos θ or 3 2 sin θ - cos θ = sin 2 θ Now, sin 2 θ = 1 - 1 - sin 2 θ = 1 - sin θ - cos θ 2 ⇒ 3 2 sin θ - cos θ = 1 - sin θ - cos θ 2 or 2 sin θ - cos θ 2 + 3 sin θ - cos θ - 2 = 0 ⇒ sin θ - cos θ + 2 { 2 sin θ - cos θ - 1 } = 0 After factorization, we get either, sin θ - cos θ + 2 = 0 . . . 1 or 2 sin θ - cos θ - 1 = 0 . . . 2 From i , cos θ - sin θ = 2 , which is not possible as max. value of cos θ - sin θ = 2 . From ii , sin θ - cos θ = 1 2 ⇒ 2 1 2 sin θ - 1 2 cos θ = 1 2 or sin θ - π 4 = 1 2 2 ⇒ θ = n π + - 1 n sin - 1 1 2 2 + π 4 , n ∈ z